Numerical Integration Questions and Answers

Q: Let \(f(x) = |x| + |x - 1| + |x - 2|\), \(x \in [-1, 2]\). Which one of the following numerical integration rules gives the exact value of the integral\(\int_{-1}^{2} f(x)\,dx\)

The exact rule is the composite trapezoidal rule by dividing [-1, 2] into 3 equal subintervals.\(f(x) = |x| + |x - 1| + |x - 2|\) is piecewise linear on [-1, 2], with corners only at \(x = 0\) and \(x = 1\). Dividing [-1, 2] into three equal parts puts the nodes at -1, 0, 1, 2, so on each subinterval \(f\) is linear and the trapezoidal rule is exact.Check: \(f(-1) = 6, f(0) = 3, f(1) = 2, f(2) = 3\), so the composite rule gives \(\frac{1}{2}(6 + 3) + \frac{1}{2}(3 + 2) + \frac{1}{2}(2 + 3) = 9.5\), the exact integral. The single trapezoid and single Simpson's rule ignore the corners, and the four-subinterval Simpson's rule places nodes at -0.25 and 1.25 with panels straddling the corners, where a quadratic cannot match a kink.

Q: A curve passing through the points as given in the table :x123456789y(x)0.20.711.31.51.71.92.12.3the area bounded by the curve, the \(x\)-axis, \(x = 1\) and \(x = 9\) is :

The area is 11.5 sq. unit.The table gives 9 equally spaced ordinates with \(h = 1\), that is 8 intervals, an even number, so Simpson's one-third rule applies:\[ \begin{aligned} A &= \tfrac{h}{3}[(y_{1} + y_{9}) + 4(y_{2} + y_{4} + y_{6} + y_{8}) \\ &\quad + 2(y_{3} + y_{5} + y_{7})] \\ &= \tfrac{1}{3}[2.5 + 4(5.8) + 2(4.4)] \\ &= \tfrac{1}{3}(34.5) = 11.5 \end{aligned} \]The trapezoidal rule gives \(1.25 + 10.2 = 11.45\), which also rounds to 11.5, so either standard rule leads to option A. 12.0, 10.5 and 10.2 are not produced by any rule; 10.2 is just the sum of the interior ordinates.

Q: Let \(y = f(x)\) takes the values \(y_{0}, y_{1}, \ldots, y_{n}\), for \(a = x_{0}\), \(x_{1} = x_{0} + h, \ldots, x_{n} = x_{0} + nh = b\). Then the value of \(\displaystyle\int_{a}^{b} f(x)dx\) by Trapezoidal rule is :

The trapezoidal rule gives \(\frac{h}{2}[(y_{0} + y_{n}) + 2(y_{1} + y_{2} + \cdots + y_{n-1})]\).Each strip \([x_{i}, x_{i+1}]\) is replaced by a trapezium of area \(\frac{h}{2}(y_{i} + y_{i+1})\). Adding the \(n\) strips, the end ordinates \(y_{0}\) and \(y_{n}\) appear once, and every interior ordinate \(y_{1}, \ldots, y_{n-1}\) appears twice, once in each neighbouring strip.The other options double the wrong ordinates: the first two or the end values instead of the interior ones.

Q: The order of magnitude of the error in estimating the value of integral \(\int_{0}^{10}(5x + 6)dx\) numerically using trapezoidal method with a step size of 0.1 is :

The error is 0.The trapezoidal rule approximates the integrand by a straight line on each sub-interval. Its error term is\[ E = -\frac{(b - a)h^{2}}{12}f''(\xi) \]which involves the second derivative. Here f(x) = 5x + 6 is itself linear, so f'' = 0 and the rule is exact apart from rounding, whatever the step size:\[ \int_{0}^{10}(5x + 6)\,dx = 250 + 60 = 310 \]The non-zero orders of magnitude in the other options would apply only to an integrand with curvature.

Q: The order of magnitude of the error in estimating the value of the integral \(\int_{0}^{1}(x^{3} + 2x^{2} + 4x + 9)\,dx\) numerically, using Simpson's 1/3rd method with a step size of 0.1 is :

The error is 0.The error term of Simpson's 1/3 rule involves the fourth derivative of the integrand:\[ E = -\frac{(b - a)h^{4}}{180}f^{(4)}(\xi) \]The integrand \(x^{3} + 2x^{2} + 4x + 9\) is a cubic, so \(f^{(4)} = 0\) and Simpson's rule is exact, whatever the step size. (Simpson's rule fits parabolas, but symmetry makes it exact for cubics too.)The powers of ten would describe the error for a function with non-zero fourth derivative, with \(h = 0.1\) giving terms of order \(h^{4} = 10^{-4}\).

Q: Let \(y = f(x)\) take the values \(y_{0}, y_{1}, \ldots, y_{n}\) for \(a = x_{0}\), \(x_{1} = x_{0} + h\), ..., \(x_{n} = x_{0} + nh = b\). Then the value of :\[ \int_{a}^{b} f(x)\, dx \]by Simpson's \(\frac{1}{3}\)-rule is :

Simpson's \(\frac{1}{3}\) rule is \(\frac{h}{3}[y_{0} + 4(y_{1} + y_{3} + \ldots + y_{n-1})\) \(+ 2(y_{2} + y_{4} + \ldots + y_{n-2}) + y_{n}]\).The rule fits a parabola through each pair of strips, giving \(\frac{h}{3}(y_{0} + 4y_{1} + y_{2})\) on the first pair. Adding over all pairs (with n even), the end ordinates appear once, the odd-numbered ordinates get weight 4, and the interior even-numbered ordinates, shared by two pairs, get weight 2.The \(\frac{h}{8}\) forms are not Simpson's rules (the \(\frac{3}{8}\) rule has factor \(\frac{3h}{8}\) and different weights). Option C puts the weight 4 on the even ordinates, including the ends, which reverses the pattern.

Q: The order of magnitude of the error in estimating the integral \(\int_{0}^{1}(x^{3} + 2x^{2} + 4x + 9)\,dx\) using Simpson's rule with step size 0.1 is :

The error is 0.The error term of Simpson's rule is proportional to the fourth derivative of the integrand:\[ E = -\frac{(b-a)h^{4}}{180} f^{(4)}(\xi) \]The integrand \(x^{3} + 2x^{2} + 4x + 9\) is a cubic, so \(f^{(4)} = 0\) everywhere and Simpson's rule gives the exact value whatever the step size. Although the rule is built by fitting parabolas, it integrates cubics exactly as well, because the cubic error cancels by symmetry on each pair of strips.The options \(10^{-3}\) to \(10^{-5}\) are the sizes one would expect from \(h^{4} = 10^{-4}\) for a general smooth function; they do not apply to a polynomial of degree three or less.

Let \(f(x) = |x| + |x - 1| + |x - 2|\), \(x \in [-1, 2]\). Which one of the following numerical integration rules gives the exact value of the integral

\(\int_{-1}^{2} f(x)\,dx\)

A.The Simpson's rule
B.The trapezoidal rule
C.The composite Simpson's rule by dividing [-1, 2] into 4 equal subintervals
D.The composite trapezoidal rule by dividing [-1, 2] into 3 equal subintervals

Exam Year(s)

GATE Mathematics - 2024,

Answer: D

The exact rule is the composite trapezoidal rule by dividing [-1, 2] into 3 equal subintervals.

\(f(x) = |x| + |x - 1| + |x - 2|\) is piecewise linear on [-1, 2], with corners only at \(x = 0\) and \(x = 1\). Dividing [-1, 2] into three equal parts puts the nodes at -1, 0, 1, 2, so on each subinterval \(f\) is linear and the trapezoidal rule is exact.

Check: \(f(-1) = 6, f(0) = 3, f(1) = 2, f(2) = 3\), so the composite rule gives \(\frac{1}{2}(6 + 3) + \frac{1}{2}(3 + 2) + \frac{1}{2}(2 + 3) = 9.5\), the exact integral. The single trapezoid and single Simpson's rule ignore the corners, and the four-subinterval Simpson's rule places nodes at -0.25 and 1.25 with panels straddling the corners, where a quadratic cannot match a kink.

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A curve passing through the points as given in the table :

x	1	2	3	4	5	6	7	8	9
y(x)	0.2	0.7	1	1.3	1.5	1.7	1.9	2.1	2.3

the area bounded by the curve, the \(x\)-axis, \(x = 1\) and \(x = 9\) is :

A.11.5 sq. unit
B.12.0 sq. unit
C.10.5 sq. unit
D.10.2 sq. unit

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The area is 11.5 sq. unit.

The table gives 9 equally spaced ordinates with \(h = 1\), that is 8 intervals, an even number, so Simpson's one-third rule applies:

\[ \begin{aligned} A &= \tfrac{h}{3}[(y_{1} + y_{9}) + 4(y_{2} + y_{4} + y_{6} + y_{8}) \\ &\quad + 2(y_{3} + y_{5} + y_{7})] \\ &= \tfrac{1}{3}[2.5 + 4(5.8) + 2(4.4)] \\ &= \tfrac{1}{3}(34.5) = 11.5 \end{aligned} \]

The trapezoidal rule gives \(1.25 + 10.2 = 11.45\), which also rounds to 11.5, so either standard rule leads to option A. 12.0, 10.5 and 10.2 are not produced by any rule; 10.2 is just the sum of the interior ordinates.

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Let \(y = f(x)\) takes the values \(y_{0}, y_{1}, \ldots, y_{n}\), for \(a = x_{0}\), \(x_{1} = x_{0} + h, \ldots, x_{n} = x_{0} + nh = b\). Then the value of \(\displaystyle\int_{a}^{b} f(x)dx\) by Trapezoidal rule is :

A.\(\frac{h}{2}[2(y_{0} + y_{1}) + (y_{2} + \cdots + y_{n})]\)
B.\(\frac{h}{2}[(y_{0} + y_{n}) + 2(y_{1} + y_{2} + \cdots + y_{n-1})]\)
C.\(\frac{h}{2}[2(y_{0} + y_{n}) + (y_{1} + y_{2} + \cdots + y_{n-1})]\)
D.\(\frac{h}{2}[(y_{0} + y_{1}) + 2(y_{2} + \cdots + y_{n})]\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The trapezoidal rule gives \(\frac{h}{2}[(y_{0} + y_{n}) + 2(y_{1} + y_{2} + \cdots + y_{n-1})]\).

Each strip \([x_{i}, x_{i+1}]\) is replaced by a trapezium of area \(\frac{h}{2}(y_{i} + y_{i+1})\). Adding the \(n\) strips, the end ordinates \(y_{0}\) and \(y_{n}\) appear once, and every interior ordinate \(y_{1}, \ldots, y_{n-1}\) appears twice, once in each neighbouring strip.

The other options double the wrong ordinates: the first two or the end values instead of the interior ones.

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The order of magnitude of the error in estimating the value of integral \(\int_{0}^{10}(5x + 6)dx\) numerically using trapezoidal method with a step size of 0.1 is :

A.0
B.\(10^{-2}\)
C.\(10^{-3}\)
D.\(10^{-1}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The error is 0.

The trapezoidal rule approximates the integrand by a straight line on each sub-interval. Its error term is

\[ E = -\frac{(b - a)h^{2}}{12}f''(\xi) \]

which involves the second derivative. Here f(x) = 5x + 6 is itself linear, so f'' = 0 and the rule is exact apart from rounding, whatever the step size:

\[ \int_{0}^{10}(5x + 6)\,dx = 250 + 60 = 310 \]

The non-zero orders of magnitude in the other options would apply only to an integrand with curvature.

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The order of magnitude of the error in estimating the value of the integral \(\int_{0}^{1}(x^{3} + 2x^{2} + 4x + 9)\,dx\) numerically, using Simpson's 1/3rd method with a step size of 0.1 is :

A.\(10^{-4}\)
B.\(10^{-5}\)
C.0
D.\(10^{-3}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The error is 0.

The error term of Simpson's 1/3 rule involves the fourth derivative of the integrand:

\[ E = -\frac{(b - a)h^{4}}{180}f^{(4)}(\xi) \]

The integrand \(x^{3} + 2x^{2} + 4x + 9\) is a cubic, so \(f^{(4)} = 0\) and Simpson's rule is exact, whatever the step size. (Simpson's rule fits parabolas, but symmetry makes it exact for cubics too.)

The powers of ten would describe the error for a function with non-zero fourth derivative, with \(h = 0.1\) giving terms of order \(h^{4} = 10^{-4}\).

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Let \(y = f(x)\) take the values \(y_{0}, y_{1}, \ldots, y_{n}\) for \(a = x_{0}\), \(x_{1} = x_{0} + h\), ..., \(x_{n} = x_{0} + nh = b\). Then the value of :

\[ \int_{a}^{b} f(x)\, dx \]

by Simpson's \(\frac{1}{3}\)-rule is :

A.\(\frac{h}{3}[y_{0} + 4(y_{1} + y_{3} + \ldots + y_{n-1})\) \(+ 2(y_{2} + y_{4} + \ldots + y_{n-2}) + y_{n}]\)
B.\(\frac{h}{8}[y_{0} + y_{n} + 4(y_{1} + y_{3} + \ldots + y_{n-1})\) \(+ 2(y_{2} + y_{4} + \ldots + y_{n-2})]\)
C.\(\frac{h}{3}[4(y_{0} + y_{2} + \ldots + y_{n})\) \(+ 2(y_{1} + y_{3} + \ldots + y_{n-1})]\)
D.\(\frac{h}{8}[4(y_{0} + y_{2} + \ldots + y_{n})\) \(+ 2(y_{1} + y_{3} + \ldots + y_{n-1})]\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

Simpson's \(\frac{1}{3}\) rule is \(\frac{h}{3}[y_{0} + 4(y_{1} + y_{3} + \ldots + y_{n-1})\) \(+ 2(y_{2} + y_{4} + \ldots + y_{n-2}) + y_{n}]\).

The rule fits a parabola through each pair of strips, giving \(\frac{h}{3}(y_{0} + 4y_{1} + y_{2})\) on the first pair. Adding over all pairs (with n even), the end ordinates appear once, the odd-numbered ordinates get weight 4, and the interior even-numbered ordinates, shared by two pairs, get weight 2.

The \(\frac{h}{8}\) forms are not Simpson's rules (the \(\frac{3}{8}\) rule has factor \(\frac{3h}{8}\) and different weights). Option C puts the weight 4 on the even ordinates, including the ends, which reverses the pattern.

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The order of magnitude of the error in estimating the integral \(\int_{0}^{1}(x^{3} + 2x^{2} + 4x + 9)\,dx\) using Simpson's rule with step size 0.1 is :

A.0
B.\(10^{-4}\)
C.\(10^{-5}\)
D.\(10^{-3}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The error is 0.

The error term of Simpson's rule is proportional to the fourth derivative of the integrand:

\[ E = -\frac{(b-a)h^{4}}{180} f^{(4)}(\xi) \]

The integrand \(x^{3} + 2x^{2} + 4x + 9\) is a cubic, so \(f^{(4)} = 0\) everywhere and Simpson's rule gives the exact value whatever the step size. Although the rule is built by fitting parabolas, it integrates cubics exactly as well, because the cubic error cancels by symmetry on each pair of strips.

The options \(10^{-3}\) to \(10^{-5}\) are the sizes one would expect from \(h^{4} = 10^{-4}\) for a general smooth function; they do not apply to a polynomial of degree three or less.

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